Question

Use completing the square to solve:
4x^2+4x-3=0

Answer 1

\[4x ^{2}+4x-3=0\]
\[x ^{2}+x-3/4=0\]
\[x ^{2}+2x(1/2)+(1/2)^{2}-(1/2)^{2}=3/4\]
\[(x+1/2)^{2}=3/4+1/4\]
\[(x+1/2)^{2}=1\]
\[x+1/2=1 or x+1/2=-1\]
\[\implies x=1/2,-3/2\] are the 2 values of x

Answer 2

so half of x is 2x?

Answer 3

\[4(x^2+x-\frac{3}{4})=0\]\[4[(x+\frac{1}{2})^2-\frac{1}{4}-\frac{3}{4}]=0\]\[(x+\frac{1}{2})^2-1=0\]\[(x+\frac{1}{2})^2=1\]\[x=\pm 1-\frac{1}{2}\]

Answer 4

@sarah you got excatly the same answer as i got

Answer 5

@chrissy who said that?

Answer 6

@luckey yeah i know i used a different method

Answer 7

dont you take half of the x term then square it??

Answer 8

@sarah both the methods takes the same time right........ i went through your answer

Answer 9

right:)

Answer 10

@chrissy thats not taking half of x and squaring it.... that was just an arrangement to make it a perfect square as;
\[(a+b)^{2}=a ^{2}+2ab+b ^{2}\]
got it??

Answer 11

sweet, thanks guys and dolls.

Answer 12

@chrissy you're always welcomed :))))

Answer 13

actually how did you get that 2x?

Answer 14

thats not 2x., thats still x only
\[2x(1/2)=x \]

Answer 15

so x(1/2)= 1/2

Answer 16

because the x is just 1

Answer 17

no no thats x(1/2) only
ok look
2x(1/2)=2(1/2)x
2 2 cancelled
then we r left with just x

Answer 18

you said before that 2x is still only x, how is that possible?

Answer 19

you please go through the solution once again properly

Answer 20

@sarah please come and explain her yaar

Answer 21

if I have £2 i have two £1 right?

Answer 22

so how can 2x remain x??? if I am being told that there are 2x's

Answer 23

the other ladies working show no 2x

Answer 24

\[x^2+x=\frac{3}{4}\] and you are now thinking that the coefficient of the "x" term is 1, as in
\[x^2+1x=\frac{3}{4}\]

Answer 25

so you think the following:
1) half of 1 is 1/2
2) one half squared is 1/4

Answer 26

yeah

Answer 27

and then write
\[(x+\frac{1}{2})^2=\frac{3}{4}+\frac{1}{4}=1\]

Answer 28

then you write
\[x+\frac{1}{2}= 1\] or
\[x+\frac{1}{2}=-1\]

Answer 29

and finally
\[x=1-\frac{1}{2}=\frac{1}{2}\] OR
\[x=-1-\frac{1}{2}=-\frac{3}{2}\]

Answer 30

if you want we can do another example

Answer 31

so what about the 2x that is not really 2x but just x business??

Answer 32

yes please another example would be great.

Answer 33

ok i do one of my choosing and write down exactly what i am thinking. start with an easy one
\[x^2+6x-1=0\]

Answer 34

add one to both sides
\[x^2+6x=1\]

Answer 35

would you mind if I tried to solve it first.

Answer 36

go ahead

Answer 37

I am stuck. i got to (x+3)=10

Answer 38

ok it goes like this
\[(x+3)^2=10\] i think you forgot the square, but the rest is right

Answer 39

then i write
\[x+3=\sqrt{10}\] or
\[x+3=-\sqrt{10}\] and finally
\[x=-3+\sqrt{10}\] or
\[x=-3-\sqrt{10}\]

Answer 40

how about
\[x^2+4x-2=0\]

Answer 41

thats is where I got stuck, 10 is not a square number so my brain froze.

Answer 42

ah ok. well we just pretend like we know what were are doing and write
\[\sqrt{10}\] if you get a perfect square, like say 4 or 9 or 16, then you know what to write

Answer 43

you can practice with
\[x^2+2x-3=0\] if you want. that will give you a perfect square

Answer 44

(x+2)^2=6
(x+2)=6???

Answer 45

you got it, but make sure if you get
\[(x+2)^2=6\] to write
\[x+2=\sqrt{6}\text { or } x+2=-\sqrt{6}\]

Answer 46

x=-3 and 1

Answer 47

for the second question

Answer 48

for the second one yes. good work! did you "complete the square"?

Answer 49

(x-1)^2

Answer 50

sorry (x+1)^2

Answer 51

got it.
here is a harder one, but still perfect square answer
\[2x^2+x-1=0\] this one is tricky so be careful

Answer 52

you where right this one is trick, the lone x term as really thrown me off.

Answer 53

still trying thought. gimme another 10 mins lol

Answer 54

yeah and what is even trickier is that you have to divide by twice! once to get the "leading coefficient " to be 1 and again when you complete the square

Answer 55

i got x = -9/4 and 7/4 :( and the square (x+1/4)^2

Answer 56

the square part is right. so you got one one side
\[(x+\frac{1}{4})^2\] and that is correct. what did you get on the other side?

Answer 57

8/16

Answer 58

ah you are off by one

Answer 59

it should be
\[x^2+\frac{1}{2}x=\frac{1}{2}\] then
\[(x+\frac{1}{4})^2=\frac{1}{2}+\frac{1}{16}\]

Answer 60

and
\[\frac{1}{16}+\frac{1}{2}=\frac{9}{16}\]

Answer 61

now when you take the square root, don't forget to take the square root of numerator and denominator, so you should have
\[x+\frac{1}{4}=\frac{3}{4}\] or
\[x+\frac{1}{4}=-\frac{3}{4}\]

Answer 62

what If i get a situation where I the numerator is a root and the denominator is not

Answer 63

like 8/9

Answer 64

you mean the other way around. no problem
\[\sqrt{\frac{8}{9}}=\frac{\sqrt{8}}{\sqrt{9}}=\frac{\sqrt{8}}{3}\]

Answer 65

that is what you usually get, because the denominator you get from squaring, so it is usually a perfect square to begin with. it is the numerator that is often not a square

Answer 66

thank you for all you help satellite

Answer 67

yw